1108. Finding Average (20)

The basic task is simple: given N real numbers, you are supposed to
calculate their average. But what makes it complicated is that some of
the input numbers might not be legal. A "legal" input is a real number
in [-1000, 1000] and is accurate up to no more than 2 decimal places.
When you calculate the average, those illegal numbers must not be
counted in.

Input Specification:

Each input file contains one test case. For each case, the first line
gives a positive integer N (<=100). Then N numbers are given in the
next line, separated by one space.

Output Specification:

For each illegal input number, print in a line "ERROR: X is not a
legal number" where X is the input. Then finally print in a line the
result: "The average of K numbers is Y" where K is the number of legal
inputs and Y is their average, accurate to 2 decimal places. In case
the average cannot be calculated, output "Undefined" instead of Y. In
case K is only 1, output "The average of 1 number is Y" instead.

这道题本来挺简单的，但是要求检查输入格式。看了一下通过率还是挺惨的。

这种题第一个想到的是正则，然而不知道为什么使用正则会导致编译超时。？_ ??

既然这样那就手工构造DFA，效率高的可怕。

还有一个办法就是利用sprintf讲读入的浮点数再比较一下。sscanf(a, "%lf", &f);sprintf(b, "%.2lf",f);

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              using namespace std;bool match_int1(char *p){ if(*p=='-') p++; int a=0;I: if(*p>='0'&&*p<='9') {p++;goto I;} if(*p==0) return true; if(*p=='.') {p++;goto F;} return false;F: if(a<2&&*p>='0'&&*p<='9') {p++,a++;goto F;} if(*p==0) return true; return false;}int main() { int n; #ifdef _DEBUG ifstream cin("input.txt");#endif cin>>n; char a[50], b[50]; double sum=0;int num=0; for(int t=0;t
              
               >a; double f; sscanf(a, "%lf", &f); if(!match_int1(a)||f<-1000||f>1000) printf("ERROR: %s is not a legal number\n",a); else { num++; sum+=f; } } if(num == 1) { printf("The average of 1 number is %.2lf", sum); } else if(num > 1) { printf("The average of %d numbers is %.2lf", num, sum / num); } else { printf("The average of 0 numbers is Undefined"); } return 0;}